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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: (x+3)*(x^2-3x+9)-(x3-9x+12)=15 (2x-1)^2+(x+3)^2-5(x+7)*(x-7)=0 25x^2-10x+11=10 (x-2)^3-(x-3)(x^2+3x+9)+6*(x+1)^2=49 ai làm nhanh em cho

Toán Lớp 8: (x+3)*(x^2-3x+9)-(x3-9x+12)=15
(2x-1)^2+(x+3)^2-5(x+7)*(x-7)=0
25x^2-10x+11=10
(x-2)^3-(x-3)(x^2+3x+9)+6*(x+1)^2=49
ai làm nhanh em cho trlhn ạ !

Comments ( 1 )

  1. \qquad (x+3)(x^2-3x+9)-(x^3-9x+12)=15
    <=> x^3+27-x^3+9x-12=15
    <=> 9x+15=15
    <=> 9x=0
    <=> x=0
    Vậy S={0}
    ————————————————–
    \qquad (2x-1)^2+(x+3)^2-5(x+7)(x-7)=0
    <=> 4x^2-4x+1+x^2+6x+9-5(x^2-49)=0
    <=> 5x^2+2x+10-5x^2+245=0
    <=> 2x+255=0
    <=> 2x=-255
    <=> x=-255/2
    Vậy S={-265/2}
    —————————————————
    \qquad 25x^2-10x+11=0
    <=> (5x)^2-2.5x.1+1+10=0
    <=> (5x-1)^2+10=0 (\text{vô lý})
    Vậy S=∅
    —————————————————
    \qquad (x-2)^3-(x-3)(x^2+3x+9)+6(x+1)^2=49
    <=> x^3-6x^2+12x-8-(x^3-27)+6(x^2+2x+1)=49
    <=> x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49
    <=> 24x+25=49
    <=> 24x=24
    <=> x=1
    Vậy S={1}
     

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222-9+11+12:2*14+14 = ? ( )